What is the principal value of \(\sec^{-1}2\)?

Correct answer: \(\frac{\pi}{3}\)

Core idea: use the identity \(\sec^{-1}x=\cos^{-1}\frac{1}{x}\) with the principal branch \([0,\pi]\) excluding \(\pi/2\). Thus \(\sec^{-1}2=\cos^{-1}\tfrac{1}{2}\). Since \(\cos\tfrac{\pi}{3}=\tfrac{1}{2}\), we get \(\cos^{-1}\tfrac{1}{2}=\tfrac{\pi}{3}\). Hence the principal value is \(\tfrac{\pi}{3}\). Closest distractor \(\tfrac{2\pi}{3}\) is incorrect because \(\sec\tfrac{2\pi}{3}=-2\), not +2. Exam tip: convert \(\sec^{-1}x\) to \(\cos^{-1}(1/x)\) and use standard cosine values (\(\pi/6,\pi/3,\pi/2\), etc.).